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3 years ago

5

if a voltage is applied to a capacitor, current flows easily at first and then slows as the capacitor becomes charged. Inductors

behave just the opposite, in that they reluctantly pass current when a voltage is first applied, and then the current passes easily as time passes. If the input voltage is suddenly raised from zero to some constant value, sketch the current in the capacitor, iC, and the inductor, iL, as a function of time. What is the steady-state current in the capacitor and inductor?

1 answer:

DENIUS [597]3 years ago

7 0

Answer: photo!

Explanation:

For the capacitor, the current as a function of time will be

I(t)=V/R e^(-t/RC)

In steady-state, time tends to infinite, so current tends to zero.

For the inductor, the current as a function of time is

I(t)=V/R (1-e^(-tR/L))

In steady-state, time tends to infinite, so current tends to V/R

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Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for

lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

<u>Length = 0.6 m = 600 mm</u>

4 0

3 years ago

Consider a multiprocessor system and a multithreaded program written using the many-to-many threading model. Let the number of u

Montano1993 [528]

Answer:

At the point when the quantity of bit strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just part strings to processors and not client level strings to processors. At the point when the quantity of part strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used all the while. Be that as it may, when a part string obstructs inside the portion (because of a page flaw or while summoning framework calls), the comparing processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, in this way expanding the use of the multiprocessor system.When the quantity of part strings is not exactly the quantity of processors, at that point a portion of the processors would stay inert since the scheduler maps just bit strings to processors and not client level strings to processors. At the point when the quantity of bit strings is actually equivalent to the quantity of processors, at that point it is conceivable that the entirety of the processors may be used at the same time. Be that as it may, when a part string hinders inside the piece (because of a page flaw or while summoning framework calls), the relating processor would stay inert. When there are more portion strings than processors, a blocked piece string could be swapped out for another bit string that is prepared to execute, along these lines expanding the usage of the multiprocessor framework.

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2 years ago

What’s is the proper fastener to use to join two wires.

xxMikexx [17]

The answer would be C !

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3 years ago

A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o

navik [9.2K]

The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

useful output energy = 2400 J

Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

h = 2400/800

h = 3 meters

Read more about Machine Efficiency at; brainly.com/question/3617034

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1 year ago

While playing a game of catch on the quadrangle, you throw a ball at an initial velocity of 17.6 m/s (approximately 39.4 mi/hr),

MAXImum [283]

Answer:

a) The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) The ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

Explanation:

a) The ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical motion at constant acceleration. First, we calculate the time taken by the ball to hit the ground:

$y = y_{o} + (v_{o}\cdot \sin \theta) \cdot t+\frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$y_{o}$ , $y$ - Initial and final vertical position, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$\theta$ - Launch angle, measured in sexagesimal degrees.

$g$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $y_{o} = 2\,m$ , $y = 0\,m$ , $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $g = -9.807\,\frac{m}{s^{2}}$ , then the time taken by the ball is:

$-4.904\cdot t^{2}+13.482\cdot t +2 = 0$ (2)

This second order polynomial can be solved by Quadratic Formula:

$t_{1} \approx 2.890\,s$ and $t_{2} \approx -0.141\,s$

Only the first root offers a solution that is physically reasonable. That is, $t \approx 2.890\,s$ .

The vertical velocity of the ball is calculated by this expression:

$v_{y} = v_{o}\cdot \sin \theta +g\cdot t$ (3)

Where:

$v_{o,y}$ , $v_{y}$ - Initial and final vertical velocity, measured in meters per second.

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ , $g = -9.807\,\frac{m}{s^{2}}$ and $t \approx 2.890\,s$ , then the final vertical velocity is:

$v_{y} = -14.860\,\frac{m}{s}$

The y-component of velocity just before the ball hits the ground is -14.860 meters per second.

b) From a) we understand that ball is in the air during approximately 2.890 seconds.

c) The horizontal distance covered by the ball ( $x$ ) is determined by the following expression:

$x = (v_{o}\cdot \cos \theta)\cdot t$ (4)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $t \approx 2.890\,s$ , then the distance covered by the ball is:

$x = 32.695\,m$

The horizontal distance covered by the ball is 32.695 meters.

d) The magnitude of the velocity of the ball just before hitting the ground ( $v$ ), measured in meters per second, is determined by the following Pythagorean identity:

$v = \sqrt{(v_{o}\cdot \cos \theta )^{2}+v_{y}^{2}}$ (5)

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , then the magnitude of the velocity of the ball is:

$v \approx 18.676\,\frac{m}{s}$ .

The magnitude of the velocity of the ball just before it hits the ground is approximately 18.676 meters per second.

e) The angle of the total velocity of the ball just before it hits the ground is defined by the following trigonometric relationship:

$\tan \theta = \frac{v_{y}}{v_{o}\cdot \cos \theta_{o}}$

If we know that $v_{o} = 17.6\,\frac{m}{s}$ , $\theta_{o} = 50^{\circ}$ and $v_{y} = -14.860\,\frac{m}{s}$ , the angle of the total velocity of the ball just before hitting the ground is:

$\theta \approx -52.717^{\circ}$

The angle of the total velocity of the ball just before it hits the ground is approximately 52.717º below the horizontal.

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2 years ago

Read 2 more answers